If dy/dx = tan (x), then y=

A) 1/2 (tan x) ^2 + C

B) (sec x) ^2 + C

C) ln|sec x| + C

D) ln|cos x| + C

E) sec x tan x + C

Question

Graham Pittman

Mathematics

2 January, 12:41

If dy/dx = tan (x), then y=

A) 1/2 (tan x) ^2 + C

B) (sec x) ^2 + C

C) ln|sec x| + C

D) ln|cos x| + C

E) sec x tan x + C

+3

Answers (2)

Know the Answer?

Reece Perkins · Answer 1

The answer to this question is:

B) (sec x) ^2 + C

Marcel Lucero · Answer 2

Dy/dx = tanx, can be answered directly using the derivatives of trigonometric functions but this is how the answer is derived

= (sinx/cosx) basic trigonometric function

= [cosx cox+sinxsinx]/cos^2x

=[cos^2x+sin^2x]/cos^2x

cos^2+sin^2x = 1; fundamental trigonometric identities

= 1/cos^2x; reciprocal relations

= sec^2x+C

The answer is letter B. sec^2x+C

If dy/dx = tan (x), then y=A) 1/2 (tan x) ^2 + CB) (sec x) ^2 + CC) ln|sec x| + CD) ln|cos x| + CE) sec x tan x + C

If dy/dx = tan (x), then y=

A) 1/2 (tan x) ^2 + C

B) (sec x) ^2 + C

C) ln|sec x| + C

D) ln|cos x| + C

E) sec x tan x + C